HDU 5019 Revenge of GCD（简单枚举）

Revenge of GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), highest common factor (hcf), or greatest common measure (gcm), of two or more integers (when at least one of them is not zero), is the largest positive integer that divides the numbers without a remainder.
---Wikipedia

Today, GCD takes revenge on you. You have to figure out the k-th GCD of X and Y.

 

Input

The first line contains a single integer T, indicating the number of test cases. 

Each test case only contains three integers X, Y and K.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= X, Y, K <= 1 000 000 000 000

 

Output

For each test case, output the k-th GCD of X and Y. If no such integer exists, output -1.

 

Sample Input

32 3 12 3 28 16 3

 

Sample Output

1-12

 

Source

BestCoder Round #10

 

题目大意：求X和Y的第K大公约数，如果不存在，输出-1。

解题思路：X 和Y的gcd一定是X和Y的第一大公约数，如果X和Y的第K大公约数存在，那么这个数一定是gcd(X,Y)的约数，因此枚举gcd(X,Y)的约数，判断是否存在第K大约数。

代码如下：

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1000000007using namespace std;typedef long long ll;typedef unsigned long long ULL;bool cmp(const ll &a,const ll &b){    return a>b;}ll gcd(ll a,ll b){    return b?gcd(b,a%b):a;}vector<ll> s;int main(){    ll i,j,a,x,y,k,t,num;    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d%I64d%I64d",&x,&y,&k);        s.clear();        a=gcd(x,y);        for(i=1;i*i<=a;i++)        {            if(a%i==0)            {                if(i*i!=a)                {                    s.push_back(i);                    s.push_back(a/i);                }                else                    s.push_back(i);            }        }        if(k>s.size())            printf("-1\n");        else        {            sort(s.begin(),s.end(),cmp);            printf("%I64d\n",s[k-1]);        }    }    return 0;}